Find Wave Function given <x>, sigmax, <p>

[friedymeister]

Homework Statement

Come up with a wave function Psi[x] that satisfies the given known values:
<x>=-1
sigma x = 1
<p> = h bar

Homework Equations

The Attempt at a Solution

So far I have this equation, which satisfies <x>, <p>, but not sigma x.
1/[Pi]^(1/4) E^(i (x + 2)) E^(-(1/2) (x + 1)^2)

 

[bdforbes]

What does sigma x refer to?

 

[friedymeister]

Sigma x is the standard deviation of x.

 

[genneth]

Remember that the standard deviation of a gaussian is controlled by dividing the x^2 term by sigma^2.

 

[yaychemistry]

Try looking at the below wikipedia article on Gaussian or Normal distributions for a good way to create a wavepacket with a given uncertainty (\sigma_x) and position \langle x\rangle.

This wikipedia link should give an adequate definition:
http://en.wikipedia.org/wiki/Normal_distribution

 

[friedymeister]

For some reason, I just can't find the function.

I don't know how I could make a wave function that could possibly have a standard distribution of 1 and an expected value of -1, graphically.

 

[yaychemistry]

A Gaussian wavefunction has the form
A\exp\left(-\frac{\left(x - x_0\right)^2}{2\sigma^2}\right)
where A is the normalization, x_0 is the center of the Gaussian, and \sigma is the standard deviation.

Try calculating
\sigma = \sqrt{\langle x^2\rangle - \langle x\rangle^2} to convince yourself of this property of Gaussians and then see if you can put the momentum part into the total wavefunction.

 

[friedymeister]

I have tried this, the only problem is that the function must also have the momentum, -h bar, which requires another exponential term, exp(-i x).

Also, sigma^2 would be 1, and that would have the form A * exp^(-i x)*exp^((-(x+1)^2)/2) where A = 1/ pi^(1/4)

Because of the first exponential function, exp^(-i x), the standard deviation becomes 1/rad(2).

How can I get around this?

 

[yaychemistry]

I'm unsure how you're getting the standard deviation to be 1/\sqrt{2} because of the complex exponential part. I'm assuming rad(2) is the square root.

Don't forget that to calculate the expectation value you use the complex conjugate on the left, e.g.
\langle x \rangle = \int \psi^{*}\left(x\right)x\psi\left(x\right)\,dx.
So the complex exponential times its complex conjugate should just give 1 inside the integral and not affect the expectation value of x and x^2.

One more thing to consider, when you multiply two Gaussians together you get another Gaussian:

A\exp\left(-\frac{\left(x-x_0\right)^2}{2\sigma^2}\right) * A\exp\left(-\frac{\left(x-x_0\right^2}{2\sigma^2}\right) = A^2\exp\left(-\frac{\left(x - x_0\right)^2}{\sigma^2}\right)
This about this relates the standard deviation of \psi\left(x\right) to \psi^{*}\left(x\right)\psi\left(x\right).

 

[friedymeister]

Yeah, that's how I've been getting <x>.

It must the be exp^(i x) that's causing the problems. The x in the exponent must be messing something up.

How can I get around this?

 

[yaychemistry]

I'm pretty sure the complex exponential should cancel in the calculation of the x and x^2 expectation values.

Remember what I mentioned about the product of Gaussians. This means that while the standard deviation of \psi\left(x\right) might be \sigma the standard deviation of |\psi\left(x\right)|^2 will be \sigma/\sqrt{2}...

 

[friedymeister]

Well, sigma x comes out to 1/rad(2) for the sqrt((integral from -inf to +inf of Psi* x^2 Psi) - (integral from -inf to +inf of Psi* x Psi)^2).

How would I change the wave function to make sigma x = 1 then?

 

[yaychemistry]

Try changing the value of sigma in the Gaussian part of the wavefunction and see how that changes what you calculate \sigma_x. See if can figure it out own your own exactly what you need to change then.
Also, remember the correct normalization of a Gaussian involves \sigma as well. It should be in the wikipedia article.

 

[friedymeister]

Oh that's awesome. Thanks, I changed 1/2 to 1/4 and then renormalized the wave function, and everything works perfectly.

Thanks a lot.

 

[yaychemistry]

Glad I could help :)